What is the time complexity of the function unifrnd(va​rMin,varMa​x,varSize)

Suppose varMin=-11, and varMax=11, and varSize=5. It means we generate a vector that has 5 elements, and these elements are bounded in [-11,11]. What is the time complexity? Constant, linear or what else?

Answer 1

[John D'Errico]

No. This does NOT generate a vector. It generates a 5x5 array as you have specified the call. (If you don't believe me, try it! Or read the help.)

If you want a vector, then you need to use varSize as [1,5] or [5,1].

The bounds of the distribution have absolutely nothing to do with the time required.

As far as the time complexity, it will be linear in the number of elements to be generated. However, for a problem as small as this, by far the most amount of time will simply be function call overhead, error checks, etc.

[Ali]

Thank you John. You are definitely right. It generates a matrix :) Sorry for misunderstanding, and thanks for the answer again.

[John D'Errico]

I'll admit that this was a design consideration that in hindsight, I think those who created some of these tools MIGHT reconsider, IF they were given the chance. I would agree that if just a scalar input is provided, too many people will assume it creates a vector with that number of elements, as opposed to a matrix.

But that aspect of the random generation functions goes back farther than I have been using MATLAB, so probably back to the very first release. (I first used MATLAB sometime in the mid to late 1980's, although I did have a copy of version 1 on disk somewhere around 1984 I'd guess.)

[John D'Errico]

A good way to test any question like this is always just to test it yourself.

timeit(@() unifrnd(-11,11,[1,1e6]))
ans =
     0.015931

timeit(@() unifrnd(-11,11,[1,1e7])) 
ans =
      0.15783

So clearly linear in the number of elements. Of course, if I make the problem too large, things will turn bad once your machine is forced to start disk swapping to work with the array. So it will only be linear in some range. Small problems will be dominated by function overhead.

timeit(@() unifrnd(-11,11,[1,5]))
ans =
   1.8231e-05

timeit(@() unifrnd(-11,11,[1,50]))
ans =
   2.1712e-05

timeit(@() unifrnd(-11,11,[1,5]))
ans =
   1.7175e-05

timeit(@() unifrnd(-11,11,[1,50]))
ans =
   1.6973e-05

So, as you can see, the estimation of the time required has a larger uncertainty than the difference between those calls for a small problem.